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    Part 1A Simple Approach

    ToShort Circuit

    Calculations

    Bulletin EDP-1(2004-1)

    EngineeringDependableProtectionFor AnElectricalDistributionSystem

    Bussmann

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    Electrical Distribution System

    Basic Considerations of Short-Circuit Calculations

    Sources of short circuit current that are normally taken

    under consideration include:

    - Utility Generation

    - Local Generation

    - Synchronous Motors and- Induction Motors

    Capacitor discharge currents can normally be

    neglected due to their short time duration. Certain IEEE

    (Institute of Electrical and Electronic Engineers) publications

    detail how to calculate these currents if they are substantial.

    Asymmetrical ComponentsShort circuit current normally takes on an asymmetrical

    characteristic during the first few cycles of duration. That is,

    it is offset about the zero axis, as indicated in Figure 1.

    Figure 1

    In Figure 2, note that the total short circuit current Ia is

    the summation of two components - the symmetrical RMS

    current IS, and the DC component, IDC. The DC componentis a function of the stored energy within the system at the

    initiation of the short circuit. It decays to zero after a few

    cycles due to I2R losses in the system, at which point the

    short circuit current is symmetrical about the zero axis. The

    RMS value of the symmetrical component may be deter-

    mined using Ohm`s Law. To determine the asymmetrical

    component, it is necessary to know the X/R ratio of the

    system. To obtain the X/R ratio, the total resistance and total

    reactance of the circuit to the point of fault must be

    determined. Maximum thermal and mechanical stress on

    the equipment occurs during these first few cycles. It is

    important to concentrate on what happens dur ing the first

    half cycle after the initiation of the fault.

    3

    Why Short-Circuit CalculationsSeveral sections of the National Electrical Code relate

    to proper overcurrent protection. Safe and reliable

    application of overcurrent protective devices based on

    these sections mandate that a short circuit study and aselective coordination study be conducted.

    These sections include, among others:

    110-9 Interrupting Rating

    110-10 Component Protection

    230-65 Service Entrance Equipment

    240-1 Conductor Protection

    250-95 Equipment Grounding Conductor Protection

    517-17 Health Care Facilities - Selective Coordination

    Compliance with these code sections can best be

    accomplished by conducting a shor t circuit study and a

    selective coordination study.

    The protection for an electrical system should not onlybe safe under all service conditions but, to insure continuity

    of service, it should be selectively coordinated as well. A

    coordinated system is one where only the faulted circuit is

    isolated without disturbing any other part of the system.

    Overcurrent protection devices should also provide short-

    circuit as well as overload protection for system

    components, such as bus, wire, motor controllers, etc.

    To obtain reliable, coordinated operation and assure

    that system components are protected from damage, it is

    necessary to first calculate the available fault current at

    various critical points in the electrical system.

    Once the short-circuit levels are determined, the

    engineer can specify proper interrupting rating require-

    ments, selectively coordinate the system and providecomponent protection.

    General Comments on Short-Circuit CalculationsShort Circuit Calculations should be done at all critical

    points in the system.

    These would include:

    - Service Entrance

    - Panel Boards

    - Motor Control Centers

    - Motor Starters

    - Transfer Switches

    - Load Centers

    Normally, short circuit studies involve calculating abolted 3-phase fault condition. This can be characterized

    as all three phases bolted together to create a zero

    impedance connection. This establishes a worst case

    condition, that results in maximum thermal and mechanical

    stress in the system. From this calculation, other types of

    fault conditions can be obtained.

    C

    U

    R

    R

    E

    N

    T

    TIME

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    3 Short-Circuit Current Calculations

    Procedures and Methods

    3 Short-Circuit Current Calculations,Procedures and Methods

    To determine the fault current at any point in the

    system, first draw a one-line diagram showing all of the

    sources of short-circuit current feeding into the fault, as well

    as the impedances of the circuit components.To begin the study, the system components, including

    those of the utility system, are represented as impedances

    in the diagram.

    The impedance tables given in the Data Section

    include three phase and single phase transformers, current

    transformers, safety switches, circuit breakers, cable, and

    busway. These tables can be used if information from the

    manufacturers is not readily available.

    It must be understood that short circuit calculations are

    performed without current limiting devices in the system.

    Calculations are done as though these devices are

    replaced with copper bars, to determine the maximum

    available short circuit current. This is necessary to

    project how the system and the current limiting devices willperform.

    Also, current limiting devices do not operate in series

    to produce a compounding current limiting effect. The

    downstream, or load side, fuse will operate alone under a

    short circuit condition if properly coordinated.

    5

    System A3 Single Transformer System

    M

    Available UtilityS.C. MVA 100,000

    1500 KVA Transformer480Y/277V,3.5%Z, 3.45%X, .56%RIf.l. = 1804A

    25 - 500kcmil6 Per PhaseService Entrance Conductorsin Steel Conduit

    2000A Switch

    KRP-C-2000SP FuseMain Swbd.

    400A Switch

    LPS-RK-400SP Fuse

    50 - 500 kcmilFeeder Cable in Steel Conduit

    Fault X2MCC No. 1

    Motor

    Fault X1

    System B3 Double Transformer System

    1000 KVA Transformer,

    480/277 Volts 33.45%X, .60%RIf.l. = 1203A

    Available UtilityS.C. KVA 500,000

    30 - 500 kcmil4 Per Phase

    Copper in PVC Conduit 1600A Switch

    KRP-C-1500SP Fuse

    LPS-RK-350SP Fuse

    400A Switch

    225 KVA208/120 Volts 3.998%X, .666%R

    20 - 2/02 Per PhaseCopper in PVC Conduit

    In this example, assume 0% motor load.

    2

    To begin the analysis, consider the following system,

    supplied by a 1500 KVA, three phase transformer having a

    full load current of 1804 amperes at 480 volts. (See System

    A, below) Also, System B, for a double transformation, will

    be studied.To start, obtain the available short-circuit KVA, MVA, or

    SCA from the local utility company.

    The utility estimates that System A can deliver a short-

    circuit of 100,000 MVA at the primary of the transformer.

    System B can deliver a short-circuit of 500,000 KVA at the

    primary of the first transformer. Since the X /R ratio of the

    utility system is usually quite high, only the reactance need

    be considered.

    With this available short-circuit information, begin to

    make the necessary calculations to determine the fault

    current at any point in the electrical system.

    Four basic methods will be presented in this text to

    instruct the reader on short circuit calculations.

    These include :- the ohmic method

    - the per unit method

    - the TRON Computer Software method

    - the point to point method

    1

    2

    Note: The above 1500KVA transformer serves 100% motor load.

    Fault X1

    Fault X2

    1

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    3 Short-Circuit Current Calculations Procedures and Methods

    Ohmic Method

    6

    3 Short Circuit Calculations,Ohmic Method

    Most circuit component impedances are given in ohms

    except utility and transformer impedances which are found

    by the following formulae* (Note that the transformer andutility ohms are referred to the secondary KV by squaring

    the secondary voltage.)

    Step 1. Xutility =1000 (KVsecondary)2

    S.C. KVAutility

    Step 2. Xtrans =(10)(%X**)(KVsecondary)2

    KVAtrans

    Rtrans =(10)(%R**)(KVsecondary)2

    KVAtrans

    Step 3. The impedance (in ohms) given for currenttransformers, large switches and large circuit breakers isessentially all X.

    Step 4. Xcable and bus .Rcable and bus.

    Step 5. Total all X and all R in system to point of fault.

    Step 6. Determine impedance (in ohms) of the system by:

    ZT = (RT)2 + (XT)2

    Step 7. Calculate short-circuit symmetrical RMS amperes

    at the point of fault.

    IS.C. sym RMS=Esecondary line-line

    3 (ZT)

    Step 8. Determine the motor load. Add up the full loadmotor currents. The full load motor current in the system is

    generally a percentage of the transformer full load current,

    depending upon the types of loads. The generally

    accepted procedure assumes 50% motor load when both

    motor and lighting loads are considered, such as supplied

    by 4 wire, 208Y/120V and 480Y/277V volt 3-phase

    systems.)

    *For simplicity of calculations all ohmic values are single phase distance one way, later compensated for in the three phase short-circuit formula by the factor, 3.(See Step 7.)

    **UL Listed transformers 25 KVA and larger have a 10% impedance tolerance. Short circuit amperes can be affected by this tolerance.Only X is considered in this procedure since utility X/R ratios are usually quite high. For more finite details obtain R of utility source.A more exact determination depends upon the sub-transient reactance of the motors in question and associated circuit impedances. A less conservativemethod would involve the total motor circuit impedance to a common bus (sometimes referred to as a zero reactance bus).

    Arithmetical addition results in conservative values of fault current. More finite values involve vectorial addition of the currents.

    Note: The ohms of the circuit components must be referred to the same voltage. If there is more than one voltage transformation in the system, the ohmicmethod becomes more complicated. It is recommended that the per-unit method be used for ease in calculation when more than one voltage transformationexists in the system.

    Step 9. The symmetrical motor contribution can beapproximated by using an average multiplying factor

    associated with the motors in the system. This factor varies

    according to motor design and in this text may be chosen

    as 4 times motor full load current for approximatecalculation purposes. To solve for the symmetrical motor

    contribution:

    Isym motor contrib = (4) x (Ifull load motor)

    Step 10. The total symmetrical short-circuit RMS current iscalculated as:

    Itotal S.C. sym RMS = (IS.C. sym RMS ) + (Isym motor contrib)

    Step 11. Determine X/R ratio of the system to the point offault.

    X/Rratio =Xtotal Rtotal

    Step 12. The asymmetrical factor corresponding to the X/Rratio in Step 11 is found in Table 8, Column Mm. This

    multiplier will provide the worst case asymmetry occurring

    in the first 1/2 cycle. When the average 3-phase multiplier

    is desired use column Ma.

    Step 13. Calculate the asymmetrical RMS short-circu itcurrent.

    IS.C. asym RMS = (IS.C. sym RMS) x (Asym Factor)

    Step 14. The short-circuit current that the motor load cancontribute is an asymmetrical current usually approximated

    as being equal to the locked rotor current of the motor.

    As a close approximation with a margin of safety use:

    Iasym motor contrib = (5) x (Ifull load motor)

    Step 15. The total asymmetrical short-circuit RMS current iscalculated as:

    Itotal S.C. asym RMS = (IS.C. asym RMS) + (Iasym motor contrib)

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    M M

    3 Short-Circuit Current Calculations Procedures and Methods

    Ohmic Method To Fault X1 System A

    Available UtilityS.C. MVA 100,000

    1500 KVA Transformer,480V, 3,3.5%Z, 3.45%X , 0.56%R

    If.l. trans = 1804A

    25 - 500 kcmil6 Per PhaseService EntranceConductors in Steel Conduit

    2000A Switch

    KRP-C-2000SP Fuse

    Fault X1

    Motor Contribution

    7

    R X

    X =1000(.48)2

    = 0.0000023 0.0000023100,000,000

    X =(10) (3.45) (.48)2

    = 0.0053 0.00531500

    R =(10) (.56) (.48)2

    = 0.00086 0.00086 1500

    X =25'

    x0.0379

    = 0.000158 0.0001581000 6

    R = 25' x 0.0244 = 0.000102 0.000102 1000 6

    X = 0.000050 0.000050

    Total R and X = 0.000962 0.00551

    Ztotal per = (0.000962)2 + (0.00551)2 = 0.0056phase

    IS.C. sym RMS =480 = 49,489A

    3 (.0056)

    Isym motor contrib = 4 x 1804 = 7216A(100% motor load)

    Itotal S.C. sym RMS = 49,489 + 7216 = 56,705A(fault X1)

    X/Rratio = .00551 = 5.73.000962

    Asym Factor = 1.294 (Table 8)

    IS.C. asym RMS = 1.294 x 49,489 = 64,039A

    Iasym motor contrib = 5 x 1804 = 9,020A(100% motor load)

    Itotal S.C. asym RMS = 64,039 + 9,020 = 73,059A(fault X1)

    One-Line D iagram Impedance D iagram

    1 1

    Note: See Ohmic Method Procedure for Formulas.

    (Table 1.2)

    (Table 3)

    (Table 5)

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    R X

    X = 0.00551 0.00551

    R = 0.000962 0.000962

    X = .00008 0.00008

    X =50 x .0379 = 0.00189 0.00189

    1000

    R =50 x .0244 = 0.00122 0.00122

    1000

    Total R and X = 0.002182 0.00748

    Ztotal per = (0.002182)2 + (0.00748)2 = 0.00778phase

    IS.C. sym RMS =480 = 35,621A

    3 (.00778)

    Isym motor contrib = 4 x 1804 = 7216A(100% motor load)

    Itotal S.C. sym RMS = 35,621 + 7,216 = 42,837A(fault X2)

    X/Rratio =.00748 = 3.43

    .002182

    Asym Factor=

    1.149 (Table 8)IS.C. asym RMS = 1.149 x 35,621 = 40,929A

    Iasym motor contrib = 5 x 1804 = 9,020A(100% motor load)

    Itotal S.C. asym RMS = 40,929 + 9,020 = 49,949A(fault X2)

    Note: See Ohmic Method Procedure for Formulas. Actual motor contributionwill be somewhat smaller than calculated due to the impedance of thefeeder cable.

    8

    3 Short-Circuit Current Calculations Procedures and Methods

    Ohmic Method To Fault X2 System A

    M M

    Adjusted Impedanceto Fault X1

    Fault X1

    400A Switch

    LPS-RK-400SP Fuse

    50 - 500 kcmilFeeder Cablein Steel Conduit

    Fault X2

    Motor Contribution

    One-Line Diagram Impedance Diagram

    2 2

    1 1

    (Table 3)

    (Table 5)

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    To use the OHMIC Method through a second transformer,

    the following steps apply:

    Step 1a. Summarize X and R values of all components onprimary side of transformer.

    One-Line Diagram Impedance Diagram

    Step 1b. Reflect X and R values of all components tosecondary side of transformer

    Xs =Vs2 (Xp) Rs =

    Vs2 (Rp)Vp2 Vp2

    and proceed with steps 2 thru 15 from page 6.

    3 Short-Circuit Current Calculations Procedures and Methods

    Ohmic Method To Fault X1 System B

    9

    R X

    X =1000 (.48)2

    = .000461 .000461500,000

    X =(10) (3.45) (.48)2

    = .00795 .007951000

    R =(10) (.60) (.48)2

    = .00138 .00138 1000

    X =30'

    x.0303 = .000227 .000227

    1000 4

    R =30'

    x.0220

    = .000165 .000165 1000 4

    X = .000050 .00005

    Total R and X = .001545 .008688

    Ztotal per = (.001545)2 + (.008688)2 = .008824phase

    IS.C. sym RMS =480 = 31,405A

    3 (.008824)

    X/Rratio =.008688 = 5.62.001545

    Asym Factor = 1.285 (Table 8)

    IS.C. asym RMS = 31,405 x 1.285 = 40,355A

    Available Utility500,000 S.C. KVA

    1000KVA Transformer,480V, 3,3.45% X, .60% R

    30' - 500 kcmil4 Per PhaseCopper in PVC Conduit

    1600A Switch

    KRP-C-1500SP Fuse

    1 1

    (Table 1.2)

    (Table 3)

    (Table 5)

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    3 Short-Circuit Current Calculations Procedures and Methods

    Ohmic Method To Fault X2 System B

    Adjusted Impedanceto fault X1

    400A Switch

    LPS-RK-350SP Fuse

    20' - 2/02 Per PhaseCopper in PVC Conduit

    225KVA Transformer,208/120V,.998%X, .666%R

    R X

    X = .008688 .008688

    R = .001545 .001545

    X = .00008 .00008

    X =20' x .0327 = .000327

    .000327

    1000 2

    R =20' x .0812 = .000812 .000812

    1000 2

    Total R and X (480V) = .002357 .009095

    To Reflect X and R to secondary:

    Xtotal =(208)2 x (.009095)

    = .001708 .001708(208V) (480)2

    Rtotal =(208)2 x (.002357)

    = .000442 .000442 (208V) (480)2

    X =(10) (.998) (.208)2

    = .00192 .00192225

    R =(10) (.666) (.208)2

    = .00128 .00128 225

    Total R and X (208V) = .001722 .003628

    Ztotal per = (.001722)2 + (.003628)2 = .004015phase

    IS.C. sym RMS =208 = 29,911A

    3 (.004015)

    X/Rratio =.003628 = 2.10.001722

    Asym Factor = 1.0491 (Table 8)

    IS.C. asym RMS = 29,911 x 1.0491 = 31,380A

    10

    One-Line Diagram Impedance Diagram

    1 1

    2 2

    (Table 5)

    (Table 1.2)

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    3 Short-Circuit Current Calculations Procedures and Methods

    Per-Unit Method

    11

    3 Short Circuit Calculation Per-Unit Method*The per-unit method is generally used for calculating

    short-circuit currents when the electrical system is more

    complex.

    After establishing a one-line diagram of the system,proceed to the following calculations: **

    Step 1. PUXutility =KVAbase

    S.C. KVAutility

    Step 2. PUXtrans =(%X)(KVAbase )

    (100)(KVAtrans)

    PURtrans =(%R)(KVAbase)

    (100)(KVAtrans)

    Step 3. PUXcomponent (cable, =(X)(KVAbase)

    switches, CT, bus) (1000)(KV)2

    Step 4. PURcomponent (cable, =(R)( KVAbase)

    switches, CT, bus) (1000)(KV)2

    Step 5. Next, total all per-unit X and all per-unit R in systemto point of fault.

    Step 6. Determine the per-unit impedance of the system by:

    PUZtotal = (PURtotal)2 + (PUXtotal)2

    Step 7. Calculate the symmetrical RMS short-circuit currentat the point of fault.

    IS.C. sym RMS=KVAbase

    3 (KV)(PUZtotal)

    Step 8. Determine the motor load. Add up the full loadmotor currents.(Whenever motor and lighting loads are

    considered, such as supplied by 4 wire, 208Y/120 and

    480Y/277 volt 3 phase systems, the generally accepted

    procedure is to assume 50% motor load based on the full

    load current rating of the transformer.)

    Step 9. The symmetrical motor contribution can beapproximated by using an average multiplying factor

    associated with the motors in the system. This factor varies

    according to motor design and in this text may be chosen

    as 4 times motor full load current for approximate

    calculation purposes. To solve for the symmetrical motorcontribution:

    ***Isym motor contrib = (4) x (Ifull load motor)

    Step 10. The total symmetrical short-circuit rms current iscalculated as:

    Itotal S.C. sym RMS = (IS.C. sym RMS) + (Isym motor contrib)

    Step 11. Determine X/R ratio of the system to the point offault.

    X/Rratio =

    PUXtotalPURtotal

    Step 12. From Table 8, Column Mm, obtain the asymmetricalfactor corresponding to the X/R ratio determined in Step

    11. This multiplier will provide the worst case asymmetry

    occurring in the first 1/2 cycle. When the average 3-phase

    multiplier is desired use column Ma.

    Step 13. The asymmetrical RMS short-circuit current can

    be calculated as:

    IS.C. asym RMS = (IS.C. sym RMS) x (Asym Factor)

    Step 14. The short-circuit current that the motor load can

    contribute is an asymmetrical current usually approximated

    as being equal to the locked rotor current of the motor.***

    As a close approximation with a margin of safety use:

    ***Iasym motor contrib = (5) x (Ifull load motor)

    Step 15. The total asymmetrical short-circuit RMS current

    is calculated as:

    ItotalS.C. asym RMS = (IS.C. asym RMS) + (Iasym motor contrib)

    * The base KVA used throughout this text will be 10,000 KVA.** As in the ohmic method procedure, all ohmic values are single-phase distance one way, later compensated for in the three phase short-circuit formula by the

    factor, 3. (See Step 7.) UL Listed transformers 25KVA and larger have a 10% impedance tolerance. Short circuit amperes can be affected by this tolerance. Only per-unit X is considered in this procedure since utility X/R ratio is usually quite high. For more finite details obtain per-unit R of utility source.

    *** A more exact determination depends upon the sub-transient reactance of the motors in question and associated circuit impedances. A less conservativemethod would involve the total motor circuit impedance to a common bus (sometimes referred to as a zero reactance bus).

    Arithmetical addition results in conservative values of fault current. More finite values involve vectorial addition of the currents.

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    *

    10,000 KVA Base

    PUR PUX

    PUX =10,000 = 0.0001 0.0001

    100,000,000

    PUX =(3.45) (10,000)

    = 0.2300 0.2300(100) (1500)

    PUR =(.56) (10,000)

    = 0.0373 0.0373 (100) (1500)

    (25')x

    (.0379)x (10,000)

    PUX =(1000) (6)

    = 0.00685 0.00685(1000) (.480)2

    (25') x (.0244) x (10,000)(1000) (6)

    PUR =(1000) (.480)2

    = 0.0044 0.0044

    PUX =(.00005) (10,000)

    = 0.00217 0.00217(1000) (.480)2

    Total PUR and PUX = 0.0417 0.2391

    PUZtotal = (0.0417)2 + (0.2391)2 = .2430

    IS.C. sym RMS =10,000 = 49,489A

    3 (.480)(.2430)

    Isym motor contrib = 4 x 1804 = 7,216A

    Itotal S.C. sym RMS = 49,489 + 7,216 = 56,705A(fault X1)

    X/Rratio =.2391 = 5.73.0417

    Asym Factor = 1.294 (Table 8)

    IS.C. asym RMS = 49,489 x 1.294 = 64,039A

    Iasym motor contrib = 5 x 1804 = 9,020A(100% motor load)

    Itotal S.C. asym RMS = 64,039 + 9,020 = 73,059A(fault X1)

    3 Short-Circuit Current Calculations Procedures and Methods

    Per-Unit Method To Fault X1 System A

    12

    Available UtilityS.C. MVA 100,000

    1500 KVA Transformer,480V, 3,3.5%Z, 3.45%X, .56%R

    If.l. trans = 1804A

    25 - 500kcmil6 Per PhaseService EntranceConductors in Steel Conduit

    2000A Switch

    KRP-C-2000SP Fuse

    MM

    Impedance DiagramOne-Line Diagram

    Note: See Per Unit Method Procedure for Formulas.Actual motor contribution will be somewhat smaller than calculateddue to impedance of the feeder cable.

    1 1

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    M M

    3 Short-Circuit Current Calculations - Procedures and Methods

    Per-Unit Method To Fault X2 System A

    13

    Fault X1

    400A Switch

    LPS-RK400SP Fuse

    50 - 500kcmilFeeder Cable inSteel Conduit

    Motor Contribution

    One-Line Diagram Impedance Diagram10,000 KVA Base

    PUR PUX

    PUX = .2391 .2391PUR = .0417 .0417

    PUX =(.00008) (10,000)

    = .0034 .0034(1000) (.480)2

    50 x (.0379) x (10,000)

    PUX =1000

    = .0822 .0822(1000) (.480)2

    50 x (.0244) x (10,000)

    PUR =1000 = .0529 .0529

    (1000) (.480)2

    Total PUR and PUX = .0946 .3247

    PUZtotal = (.0946)2

    + (.3247)2

    = 0.3380

    IS.C. sym RMS =10,000 = 35,621A

    3 (.480)(.3380)

    Isym motor contrib = 4 x 1804 = 7,216A

    Itotal S.C. sym RMS = 35,621 + 7,216 = 42,837A(fault X2)

    X/Rratio =.32477

    = 3.43.09465

    Asym Factor = 1.149 (Table 8)

    IS.C. asym RMS = 1.149 x 35,621 = 40,929A

    Iasym motor contrib = 5 x 1804 = 9,020A(100%motor load)

    Itotal S.C. asym RMS = 40,929 + 9,020 = 49,949A(fault X2)

    Adjusted Impedanceto Fault X

    1

    1 1

    2 2

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    10,000KVA BasePUR PUX

    PUX =10,000 = .02 .02

    500,000

    PUX =(3.45) (10,000)

    = .345 .345(100) (1000)

    PUR =(.6) (10,000)

    = .06 .06 (100) (1000)

    (30')x

    (.0303)

    PUX =(1000) (4)

    x (10,000)

    = .0099 .0099(1000) (.48)2

    (30') x (.0220)

    PUR =(1000) (4)

    x (10,000)

    = .0072 .0072 (1000) (.48)2

    PUX =(.00005) (10,000)

    = .0022 .0022(1000) (.48)2

    Total PUR and PUX = .0672 .3771

    PUZtotal

    = (.0672)2 + (.3771)2 = .383

    IS.C. sym RMS =10,000 = 31,405A

    3 (.48)(.383)

    X/Rratio =.3771 = 5.62.0672

    Asym Factor = 1.285 (Table 8)

    IS.C.asym RMS = 31,405 x 1.285 = 40,355A

    14

    3 Short-Circuit Current Calculations Procedures and Methods

    Per-Unit Method To Fault X1 System B

    Available UtilityS.C. KVA 500,000

    1000 KVA Transformer,480V, 33.45%X, .60%R

    30' - 500kcmil

    4 Per PhaseCopper in PVC Conduit

    1600A Switch

    KRP-C-1500SP Fuse

    One-Line Diagram Impedance Diagram

    1 1

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    3 Short-Circuit Current Calculations Procedures and Methods

    Per-Unit Method To Fault X2 System B

    15

    One-Line Diagram Impedance Diagram10,000 KVA

    PUR PUX

    X1 = .3771 .3771

    R1 = .0672 .0672

    PUX =(.00008) (10,000)

    = .0035 .0035(1000) (.48)2

    PUX =

    (20')x

    (.0327)x (10,000)

    = .0142 .0142(1000) (2)

    (1000) (.48)2

    (20')x

    (.0812)x (10,000)

    PUR =(1000) (2)

    = .0352 .0352 (1000) (.48)2

    PUX =(.998) (10,000)

    = .4435 .4435(100) (225)

    PUR =(.666) (10,000)

    = .296 .296 (100) (225)

    Total PUR and PUX .3984 .8383

    PUZtotal = (.3984)2 + (.8383)2 = .928

    IS.C.sym RMS = 10,000 = 29,911A(3)(.208)(.928)

    X/Rratio =.8383 = 2.10.3984

    Asym Factor=

    1.0491 (Table 8)IS.C. asym RMS = 29,911 x 1.0491 = 31,380A

    Adjusted Impedance toFault X1

    400A Switch

    LPS-RK-350SP Fuse

    20 - 2/02 Per PhaseCopper in PVC conduit

    225KVA Transformer,208V, 3.998%X, .666%R

    2 2

    1 1

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    BUSSPOWER is a Computer Software Program which

    calculates three phase fault currents. It is a part of the

    TRON Software Package for Power Systems Analysis.

    The user inputs data which includes:

    - Cable and Busway Lengths and Types- Transformer Rating and Impedence- Fault sources such as Utility Available and Motor

    Contribution.

    Following the data input phase, the program is

    executed and an output report reviewed.

    The following is a partial output report of System A

    being studied.

    TRON Software Fault Calculation Program Three Phase Fault Report

    SYSTEM AFault Study Summary

    Bus Record Voltage Available RMS Duties

    Name L-L 3 Phase Momentary

    (Sym) (Asym)

    X1 480 58414 77308

    X2 480 44847 53111

    The following is a partial output report of the

    distribution System B.

    SYSTEM BFault Study Summary

    Bus Record Voltage Available RMS Duties

    Name L-L 3 Phase Momentary

    (Sym) (Asym)X1 480 31,363 40,141

    X2 208 29,980 31,425

    A further description of this program and its

    capabilities is on the back cover of this bulletin.

    3 Short-Circuit Current Calculations Procedures and Methods

    TRON Computer Software Method

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    The application of the point-to-point method permits the

    determination of available short-circuit currents with a

    reasonable degree of accuracy at various points for either

    3 or 1 electrical distribution systems. This method can

    assume unlimited primary short-circuit current (infinite bus).

    Basic Point-to-Point Calculation ProcedureStep 1. Determine the transformer full load amperes fromeither the nameplate or the following formulas:

    3 Transformer If.l. =KVA x 1000EL-L x 1.732

    1 Transformer If.l. =KVA x 1000

    EL-L

    Step 2. Find the transformer multiplier.

    Multiplier = 100*%Z trans

    Note. Transformer impedance (Z) helps to determine what the short circuitcurrent will be at the transformer secondary. Transformer impedance is

    determined as follows: The transformer secondary is short circuited. Voltageis applied to the primary which causes full load current to flow in thesecondary. This applied voltage divided by the rated primary voltage is theimpedance of the transformer.Example: For a 480 volt rated primary, if 9.6 volts causes secondary full loadcurrent to flow through the shorted secondary, the transformer impedance is9.6/480 = .02 = 2%Z.In addition, UL listed transformer 25KVA and larger have a 10%impedance tolerance. Short circuit amperes can be affected by thistolerance.

    Step 3. Determine the transformer let-thru short-circuitcurrent**.

    IS.C. = If.l. x Multiplier

    Note. Motor short-circuit contribution, if significant, may be added to the

    transformer secondary short-circuit current value as determined in Step 3.Proceed with this adjusted figure through Steps 4, 5 and 6. A practicalestimate of motor short-circuit contribution is to multiply the total motorcurrent in amperes by 4.

    Step 4. Calculate the "f" factor.

    3 Faults f = 1.732 x L x IC x EL-L

    1 Line-to-Line (L-L)2 x L x I

    Faults on 1 Center f =C x EL-LTapped Transformer

    1 Line-to-Neutral2 x L x I(L-N) Faults on 1 f =C x EL-NCenter Tapped Transformer

    Where:L = length (feet) of circuit to the fault.C = constant from Table 6, page 27. For parallel

    runs, multiply C values by the number of

    conductors per phase.

    I = available short-circuit current in amperes atbeginning of circuit.

    Note. The L-N fault current is higher than the L-L fault current at thesecondary terminals of a single-phase center-tapped transformer. Theshort-circuit current available (I) for this case in Step 4 should be adjustedat the transformer terminals as follows:At L-N center tapped transformer terminals,I = 1.5 x L-L Short-Circuit Amperes at Transformer Terminals

    At some distance from the terminals, depending upon wire size, the L-N faultcurrent is lower than the L-L fault current. The 1.5 multiplier is anapproximation and will theoretically vary from 1.33 to 1.67. These figures arebased on change in turns ratio between primary and secondary, infinitesource available, zero feet from terminals of transformer, and 1.2 x %X and1.5 x %R for L-N vs. L-L resistance and reactance values. Begin L-Ncalculations at transformer secondary terminals, then proceed point-to-point.

    Step 5. Calculate "M" (multiplier).

    M = 11 + f

    Step 6. Calculate the available short-circuit symmetricalRMS current at the point of fault.

    IS.C. sym RMS = IS.C. x M

    Calculation of Short-Circuit Currentsat Second Transformer in System

    Use the following procedure to calculate the level of

    fault current at the secondary of a second, downstream

    transformer in a system when the level of fault current at the

    transformer primary is known.

    Procedure for Second Transformer in System

    Step 1. Calculate the "f" factor (IS.C. primary known)

    3 Transformer(IS.C. primary and f =

    IS.C. primary x Vprimary x 1.73 (%Z)

    IS.C. secondary are 100,000 x KVA trans3 fault values)

    1 Transformer(IS.C. primary andIS.C. secondary are f =

    IS.C. primary x Vprimary x (%Z)

    1 fault values: 100,000x KVA trans

    IS.C. secondary is L-L)

    Step 2. Calculate "M" (multiplier).

    M =1

    1 + f

    Step 3. Calculate the short-circuit current at the secondaryof the transformer. (See Note under Step 3 of "Basic Point-

    to-Point Calculation Procedure".)

    IS.C. secondary =Vprimary x M x IS.C. primary

    Vsecondary

    **

    *

    3 Short-Circuit Current Calculations Procedures and Methods

    Point-to-Point Method

    17

    IS.C. primary

    MAINTRANSFORMER

    H.V. UTILITYCONNECTION

    IS.C. secondary

    IS.C. primary IS.C. secondary

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    Available UtilityS.C. MVA 100,000

    1500 KVA Transformer,480V, 3, 3.5%Z,3.45%X, 56%R

    If.l. =1804A

    25' - 500kcmil6 Per PhaseService EntranceConductors in Steel Conduit

    2000A Switch

    KRP-C-2000SP Fuse

    Fault X1

    400A Switch

    LPS-RK-400SP Fuse

    50' - 500 kcmilFeeder Cablein Steel Conduit

    Fault X2

    Motor Contribution M

    3 Short-Circuit Current Calculations Procedures and Methods

    Point-to-Point Method To Faults X1 & X2 System A

    Fault X1

    Step 1. If.l. =1500 x 1000 = 1804A480 x 1.732

    Step 2. Multiplier = 100 = 28.573.5

    Step 3. IS.C.= 1804 x 28.57 = 51,540A

    Step 4. f =1.732 x 25 x 51,540 = 0.0349

    6 x 22,185 x 480

    Step 5. M =1 = .9663

    1 + .0349

    Step 6. IS.C.sym RMS = 51,540 x .9663 = 49,803A

    IS.C.motor contrib = 4 x 1,804 = 7,216A

    ItotalS.C. sym RMS = 49,803 + 7,216 = 57,019A(fault X1)

    Fault X2

    Step 4. Use IS.C.sym RMS @ Fault X1 to calculate "f"

    f =1.732 x 50 x 49,803 = .4050

    22,185 x 480

    Step 5. M =1 = .7117

    1 + .4050

    Step 6. IS.C.sym RMS = 49,803 x .7117 = 35,445A

    Isym motor contrib = 4 x 1,804 = 7,216A

    Itotal S.C. sym RMS(fault X2)

    = 35,445 + 7,216 = 42,661A

    1

    2

    18

    One-Line Diagram

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    3 Short-Circuit Current Calculations RMS Amperes

    Comparison of Results

    Fault X1

    Step 1. If.l. =1000 x 1000 = 1203A480 x 1.732

    Step 2. Multiplier = 100 = 28.573.5

    Step 3. IS.C. = 1203 x 28.57 = 34,370A

    Step 4. f =1.732 x 30 x 34,370 = .0348

    4 x 26,706 x 480

    Step 5. M =1 = .9664

    1 + .0348

    Step 6. IS.C.sym RMS = 34,370 x .9664 = 33,215A

    Fault X2

    Step 4. f = 1.732 x 20 x 33,215 = .10492 x 11,423 x 480

    Step 5. M = 1 = .9051 + .1049

    Step 6. IS.C.sym RMS = 33,215 x .905 = 30,059A

    Fault X2

    f =30,059 x 480 x 1.732 x 1.2 = 1.333

    100,000 x 225

    M =1 = .4286

    1 + 1.333

    IS.C. sym RMS =480 x .4286 x 30,059 = 29,731A

    208

    3 Short-Circuit Current Calculations Procedures and Methods

    Point-to-Point Method To Faults X1 & X2 - System B

    Available Utility500,000 S.C KVA

    1000 KVA Transformer,480V, 3,3.5%Z

    If.l.= 1203A

    30 - 500 kcmil4 Per PhaseCopper in PVC Conduit

    1600A Switch

    KRP-C-1500SP Fuse

    Fault X1

    400A Switch

    LPS-RK-350SP Fuse

    20 - 2/02 Per PhaseCopper in PVC Conduit

    225 KVA transformer,208V, 31.2%Z

    One-Line Diagram

    1

    2

    19

    System A

    Ohmic Per-Unit TRON PTPSym. Asym. Sym. Asym. Sym. Asym. Sym.

    X1W/O Motor 49,489 64,039 49,489 64,039 49,992 64,430 49,803

    W/Motor 56,705 73,059 56,705 73,059 58,414 77,308 57,019

    X2W/O Motor 35,621 40,929 35,621 40,929 36,126 41,349 35,445

    W/Motor 42,837 49,949 42,837 49,949 44,847 53,111 42,661

    Notes:1. OHMIC and PER UNIT methods assume 100% motor contribution at X1,

    then at X2.2. TRON modeled 100% motor contribution by assuming 1500 HP load,

    located at Point X2.3. PTP method added symmetrical motor contribution at X1, then at X2.

    System B

    Ohmic Per-Unit TRON PTPSym. Asym. Sym. Asym. Sym. Asym. Sym.

    X1 31,405 40,355 31,405 40,355 31,363 40,145 33,215

    X2 29,911 31,380 29,911 31,380 29,980 31,425 29,731

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    1 Short-Circuit Current Calculations 1 Transformer System

    Procedures and Methods

    Short-circuit calculations on a single-phase center

    tapped transformer system require a slightly different

    procedure than 3 faults on 3 systems.

    1. It is necessary that the proper impedance be used to

    represent the primary system. For 3 fault calculations, asingle primary conductor impedance is only considered

    from the source to the transformer connection. This is

    compensated for in the 3 short-circuit formula by

    multiplying the single conductor or single-phase

    impedance by 1.73.

    However, for single-phase faults, a primary conductor

    impedance is considered from the source to the

    transformer and back to the source. This is compensated in

    the calculations by multiplying the 3 primary source

    impedance by two.

    2. The impedance of the center-tapped transformer must

    be adjusted for the half-winding (generally line-to-neutral)fault condition.

    The diagram at the right illustrates that during line-to-

    neutral faults, the full primary winding is involved but, only

    the half-winding on the secondary is involved. Therefore,

    the actual transformer reactance and resistance of the half-

    winding condition is different than the actual transformer

    reactance and resistance of the full winding condition.

    Thus, adjustment to the %X and %R must be made when

    considering line-to-neutral faults. The adjustment multipliers

    generally used for this condition are as follows:

    1.5 times full winding %R on full winding basis.

    1.2 times full winding %X on full winding basis.

    Note: %R and %X multipliers given in Table 1.3 may be used, however,calculatios must be adjusted to indicate transformer KVA/2.

    3. The impedance of the cable and two-pole switches on

    the system must be considered "both-ways" since the

    current flows to the fault and then returns to the source. For

    instance, if a line-to-line fault occurs 50 feet from a

    transformer, then 100 feet of cable impedance must be

    included in the calculation.

    The calculations on the following pages illustrate 1

    fault calculations on a single-phase transformer system.

    Both line-to-line and line-to-neutral faults are considered.

    Note in these examples:

    a. The multiplier of 2 for some electrical components to

    account for the single-phase fault current flow,

    b. The half-winding transformer %X and %R multipliers for

    the line-to-neutral fault situation,and

    c. The KVA and voltage bases used in the per-unit

    calculations

    PRIMARY

    SECONDARY

    SHORTCIRCUIT

    PRIMARY

    SECONDARY

    SHORT CIRCUIT

    L2 N L1

    SHORT CIRCUIT

    L1

    N

    L2

    A

    B

    C

    50 feet

    20

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    100,000 KVA3 Source

    75KVA, 1 Transformer,1.22%X, .68%R

    Negligible Distance

    400A Switch

    LPN-RK-400SP Fuse

    25' - 500kcmil

    Magnetic Conduit

    1 Short-Circuit Current Calculations 1 Transformer System

    Per-Unit Method Line-to-Line Fault @ 240V Fault X1

    10,000KVA Base

    PUR PUX

    PUX(3) =10,000 = .1100,000

    PUX(1) = 2 x .1 = .2000 .2000

    PUX =(1.22) (10,000)

    = 1.6267 1.6267(100) (75)

    PUR =(.68) (10,000)

    = .9067 .9067 (100) (75)

    PUX =2(.00008) (10,000)

    = .0278 .0278(1000) (.240)2

    25'

    PUX =2 x

    1000x .0379 x 10,000

    = .3289 .3289(1000) (.240)2

    25'

    PUR=2 x

    1000x .0244 x 10,000

    = .2118 .2118 (1000) (.240)2

    Total PUR and PUX = 1.1185 2.1834

    PUZtotal = (1.1185)2 + (2.1834)2 = 2.4532

    IS.C. sym RMS =10,000

    = 16,984AL-L @ 240V (.240) (2.4532)

    Note: See "Data Section" for impedance data for the electrical components.

    One-Line Diagram

    1 1

    21

    Impedance Diagram

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    *

    **

    1 Short-Circuit Current Calculations 1 Transformer System

    Per-Unit Method Line-to-Neutral Fault @ 120V Fault X1

    100,000 KVA3 Source

    75KVA, 1 Transformer,1.22%X, .68%R

    Negligible Distance

    400A Switch

    LPN-RK-400SP Fuse

    25' - 500kcmil

    Magnetic Conduit

    10,000KVA Base

    PUR PUX

    PUX(3) =10,000 = .1100,000

    PUX(1) = 2 x .1 = .2000 .2000

    PUX =(1.2) (1.22) (10,000)

    = 1.952 1.952(100) (75)

    PUR =(1.5) (.68) (10,000)

    = 1.3600 1.3600 (100) (75)

    PUX* =(.00008) (10,000)

    = .0556 .0556(1000) (.120)2

    25'

    PUX** =2 x

    1000x .0379 x 10,000

    = 1.316 1.316(1000) (.120)2

    25'

    PUR** =2 x

    1000x .0244 x 10,000

    = .8472 .8472 (1000) (.120)2

    Total PUR and PUX = 2.2072 3.5236

    PUZtotal = (2.2072)2 + (3.5236)2 = 4.158

    IS.C. sym RMS =10,000

    = 20,041AL-N @ 120V (.120) (4.158)

    Note: See "Data Section" for impedance data for the electrical components.

    The multiplier of two (2) is not applicable since on a line to neutral fault, onlyone switch pole is involved.

    Assumes the neutral conductor and the line conductor are the same size.

    1 1

    One-Line Diagram Impedance Diagram

    22

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    1 Short-Circuit Current Calculations 1 Transformer System

    Point-to-Point Method Line-to-Line Fault @ 240V Fault X1

    Available UtilityS.C.KVA 100,0003 Source

    75KVA, 1 Transformer,1.22%X, .68%R1.40%Z120/240V

    Negligible Distance

    400A Switch

    LPN-RK-400SP Fuse

    25' - 500kcmil

    Magnetic Conduit

    One-Line Diagram

    1

    Fault X1

    Step 1. If.l. =75 x 1000 = 312.5A

    240

    Step 2. Multiplier =100 = 71.431.40

    Step 3. IS.C. = 312.5 x 71.43 = 22,322A

    Step 4. f = 2x 25 x 22,322 = .2096

    22,185 x 240

    Step 5.1 = .8267M =

    1 + .2096

    Step 6. IS.C. L-L (X1) = 22,322 x .8267 = 18,453A

    23

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    *

    1 Short Circuit Calculations RMS Amperes

    Comparison of Results

    Per-Unit Method vs. Point-to-Point MethodPer-Unit PTP

    Method Method

    X1Line-Line 16,984A 18,453A

    Line-Neutral 20,041A 20,555A

    Fault X1

    Step 1. If.l. =75 x 1000 = 312.5A

    240

    Step 2. Multiplier =100 = 71.431.40

    Step 3. IS.C. (L-L) = 312.5 x 71.43 = 22,322A

    IS.C. (L-N) = 22,322 x 1.5 = 33,483A

    Step 4. f =2*x 25 x 22,322 x 1.5

    = .628822,185 x 120

    Step 5. 1 = .6139M =1 + .6288

    Step 6. IS.C. L-N (X1) = 33,483 x .6139 = 20,555A

    Assumes the Neutral conductor and the line conductor are the same size.

    Available UtilityS.C.KVA 100,0003 Source

    75KVA, 1 Transformer,1.22% X, .68%R,1.40%Z120/240V

    Negligible Distance

    400A Switch

    LPN-RK-400SP Fuse

    25' - 500kcmil

    Magnetic Conduit

    One-Line Diagram

    1 Short-Circuit Current Calculations 1 Transformer System

    Point-to-Point Method Line-to-Neutral Fault @ 120V Fault X1

    1

    24

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    *

    **

    Table 1.4. Impedance Data for Single Phase and Three PhaseTransformers-Supplement

    KVA Suggested

    1 3 %Z X/R Ratio for Calculation

    10 1.2 1.1

    15 1.3 1.175 1.11 1.5

    150 1.07 1.5

    225 1.12 1.5

    300 1.11 1.5

    333 1.9 4.7

    500 2.1 5.5

    These represent actual transformer nameplate ratings taken from field

    installations.

    Note: UL Listed transformers 25KVA and greater have a 10% tolerance on

    their impedance nameplate.

    Table 2. Current Transformer Reactance DataApproximate Reactance of Current Transformers*

    Reactance in Ohms for

    Primary Current Various Voltage Ratings

    Ratings - Amperes 600-5000V 7500V 15,000V

    100 - 200 0.0022 0.0040

    250 - 400 0.0005 0.0008 0.0002

    500 - 800 0.00019 0.00031 0.00007

    1000 - 4000 0.00007 0.00007 0.00007

    Note: Values given are in ohms per phase. For actual values, refer to manu-

    facturers' data.

    This table has been reprinted from IEEE Std 241-1990, IEEE Recommended

    Practice for Commercial Building Power Systems, Copyright 1990 by the

    Institute of Electrical and Electronics Engineers, Inc. with the permission of

    the IEEE Standards Department.

    Table 3. Disconnecting Switch Reactance Data(Disconnecting-Switch Approximate Reactance Data, in Ohms*)Switch Size Reactance(Amperes) (Ohms)

    200 0.0001400 0.00008

    600 0.00008

    800 0.00007

    1200 0.00007

    1600 0.00005

    2000 0.00005

    3000 0.00004

    4000 0.00004

    Note: The reactance of disconnecting switches for low-voltage circuits

    (600V and below) is in the order of magnitude of 0.00008 - 0.00005

    ohm/pole at 60 Hz for switches rated 400 - 4000 A, respectively.

    *For actual values, refer to manufacturers data.

    This table has been reprinted from IEEE Std 241-1990, IEEE Recommended

    Practice for Commercial Building Power Systems, Copyright 1990 by the

    Institute of Electrical and Electronics Engineers, Inc. with the permission ofthe IEEE Standards Department.

    1 Pole

    25

    Table 1.1. Transformer Impedance Data(X/R Ratio of Transformers Based on ANSI/IEEE C37.010-1979)

    This table has been reprinted from IEEE Std 141-1986, IEEE Recommended

    Practice for Electric Power Distribution for Industrial Plants, Copyright 1986

    by the Institute of Electrical and Electronics Engineers, Inc with the

    permission of the IEEE Standards Department.

    Table 1.2. Impedance Data for Three Phase TransformersKVA %R %X %Z X/R

    3.0 3.7600 1.0000 3.8907 0.265

    6.0 2.7200 1.7200 3.2182 0.632

    9.0 2.3100 1.1600 2.5849 0.502

    15.0 2.1000 1.8200 2.7789 0.867

    30.0 0.8876 1.3312 1.6000 1.5

    45.0 0.9429 1.4145 1.7000 1.5

    75.0 0.8876 1.3312 1.6000 1.5

    112.5 0.5547 0.8321 1.0000 1.5

    150.0 0.6657 0.9985 1.2000 1.5

    225.0 0.6657 0.9985 1.2000 1.5

    300.0 0.6657 0.9985 1.2000 1.5

    500.0 0.7211 1.0816 1.3000 1.5

    750.0 0.6317 3.4425 3.5000 5.451000.0 0.6048 3.4474 3.5000 5.70

    1500.0 0.5617 3.4546 3.5000 6.15

    2000.0 0.7457 4.9441 5.0000 6.63

    2500.0 0.7457 4.9441 5.0000 6.63

    Note: UL Listed transformers 25KVA and greater have a 10% tolerance on

    their nameplate impedance.

    Table 1.3. Impedance Data for Single Phase TransformersSuggested Normal Range Impedance Multipliers**

    X/R Ratio of Percent For Line-to-Neutral

    kVA for Impedance (%Z)* Faults

    1 Calculation for %X for%R

    25.0 1.1 1.26.0 0.6 0.75

    37.5 1.4 1.26.5 0.6 0.75

    50.0 1.6 1.26.4 0.6 0.75

    75.0 1.8 1.26.6 0.6 0.75

    100.0 2.0 1.35.7 0.6 0.75

    167.0 2.5 1.46.1 1.0 0.75

    250.0 3.6 1.96.8 1.0 0.75

    333.0 4.7 2.46.0 1.0 0.75

    500.0 5.5 2.25.4 1.0 0.75

    National standards do not speciify %Z for single-phase transformers. Consultmanufacturer for values to use in calculation.Based on rated current of the winding (onehalf nameplate kVA divided bysecondary line-to-neutral voltage).

    Note: UL Listed transformers 25 KVA and greater have a 10% tolerance ontheir impedance nameplate.

    This table has been reprinted from IEEEStd 242-1986 (R1991), IEEE

    Recommended Practice for Protection and Coordination of Industrial and

    Commercial Power Systems, Copyright 1986 by the Institute of Electrical

    and Electronics Engineers, Inc. with the permission of the IEEE Standards

    Department.

    50

    40

    30

    20

    10

    0

    TypicalX/R

    0.5 1 2 5 10 20 50 100 200 500 1000Self-Cooled Transformer Rating in MVA

    Data Section

    Impedance and Reactance DataTransformers and Switches

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    Data Section

    Impedance & Reactance Data-Circuit Breakers and Conductors

    Table 5. Impedance Data - Insulated Conductors(Ohms/1000 ft. each conductor - 60Hz)Size Resistance (25C) Reactance - 600V - THHN

    AWG or Copper Aluminum Single Conductors 1 Multiconductor

    kcM Metal NonMet Metal Nonmet Mag. Nonmag. Mag Nonmag.

    14 2.5700 2.5700 4.2200 4.2200 .0493 .0394 .0351 .030512 1.6200 1.6200 2.6600 2.6600 .0468 .0374 .0333 .0290

    10 1.0180 1.0180 1.6700 1.6700 .0463 .0371 .0337 .0293

    8 .6404 .6404 1.0500 1.0500 .0475 .0380 .0351 .0305

    6 .4100 .4100 .6740 .6740 .0437 .0349 .0324 .0282

    4 .2590 .2590 .4240 .4240 .0441 .0353 .0328 .0235

    2 .1640 .1620 .2660 .2660 .0420 .0336 .0313 .0273

    1 .1303 .1290 .2110 .2110 .0427 .0342 .0319 .0277

    1/0 .1040 .1020 .1680 .1680 .0417 .0334 .0312 .0272

    2/0 .0835 .0812 .1330 .1330 .0409 .0327 .0306 .0266

    3/0 .0668 .0643 .1060 .1050 .0400 .0320 .0300 .0261

    4/0 .0534 .0511 .0844 .0838 .0393 .0314 .0295 .0257

    250 .0457 .0433 .0722 .0709 .0399 .0319 .0299 .0261

    300 .0385 .0362 .0602 .0592 .0393 .0314 .0295 .0257

    350 .0333 .0311 .0520 .0507 .0383 .0311 .0290 .0254

    400 .0297 .0273 .0460 .0444 .0385 .0308 .0286 .0252

    500 .0244 .0220 .0375 .0356 .0379 .0303 .0279 .0249

    600 .0209 .0185 .0319 .0298 .0382 .0305 .0278 .0250750 .0174 .0185 .0264 .0240 .0376 .0301 .0271 .0247

    1000 .0140 .0115 .0211 .0182 .0370 .0296 .0260 .0243

    Note: Increased resistance of conductors in magnetic raceway is due to the effect of hysteresislosses. The increased resistance of conductors in metal non-magnetic raceway is due to the effectof eddy current losses. The effect is essentially equal for steel and aluminum raceway. Resistancevalues are acceptable for 600 volt, 5KV and 15 KV insulated Conductors.

    Size Reactance - 5KV Reactance - 15KV

    AWG or Single Conductors 1 Multiconductor Single Conductors 1 Multiconductor

    kcM Mag. Nonmag. Mag. Nonmag. Mag. Nonmag. Mag. Nonmag.

    8 .0733 .0586 .0479 .0417

    6 .0681 .0545 .0447 .0389 .0842 .0674 .0584 .0508

    4 .0633 .0507 .0418 .0364 .0783 .0626 .0543 .0472

    2 .0591 .0472 .0393 .0364 .0727 .0582 .0505 .0439

    1 .0571 .0457 .0382 .0332 .0701 .0561 .0487 .0424

    1/0 .0537 .0430 .0360 .0313 .0701 .0561 .0487 .0424

    2/0 .0539 .0431 .0350 .0305 .0661 .0561 .0458 .0399

    3/0 .0521 .0417 .0341 .0297 .0614 .0529 .0427 .03724/0 .0505 .0404 .0333 .0290 .0592 .0491 .0413 .0359

    250 .0490 .0392 .0323 .0282 .0573 .0474 .0400 .0348

    300 .0478 .0383 .0317 .0277 .0557 .0458 .0387 .0339

    350 .0469 .0375 .0312 .0274 .0544 .0446 .0379 .0332

    400 .0461 .0369 .0308 .0270 .0534 .0436 .0371 .0326

    500 .0461 .0369 . 0308 .0270 .0517 .0414 .0357 .0317

    600 .0439 .0351 .0296 .0261 .0516 .0414 . 0343 .0309

    750 .0434 .0347 .0284 .0260 .0500 .0413 .0328 .0301

    1000 .0421 .0337 .0272 .0255 .0487 .0385 .0311 .0291

    These are only representative figures. Reactance is affected by cable insulation type, shielding,

    conductor outside diameter, conductor spacing in 3 conductor cable, etc. In commercial buildings

    meduim voltage impedances normally do not affect the short circuit calculations significantly.

    This table has been reprinted from IEEE Std 241-1990, IEEE Recommended Practice for Commercial

    Building Power Systems, copyright 1990 by the Institute of Electrical and Electronics Engineers,

    Inc. with the permission of the IEEE Standards Department.

    *

    Table 4. Circuit Breaker Reactance Data(a) Reactance of Low-Voltage Power Circuit Breakers

    Circuit-Breaker

    Interrupting Circuit-Breaker

    Rating Rating Reactance(amperes) (amperes) (ohms)

    15,000 15 - 35 0.04

    and 50 - 100 0.004

    25,000 125 - 225 0.001

    250 - 600 0.0002

    50,000 200 - 800 0.0002

    1000 - 1600 0.00007

    75,000 2000 - 3000 0.00008

    100,000 4000 0.00008

    (b)Typical Molded Case Circuit Breaker Impedances

    Molded-Case

    Circuit-Breaker

    Rating Resistance Reactance

    (amperes) (ohms) (ohms)

    20 0.00700 Negligible

    40 0.00240 Negligible

    100 0.00200 0.00070

    225 0.00035 0.00020

    400 0.00031 0.00039

    600 0.00007 0.00017

    Notes:

    (1) Due to the method of rating low-voltage powercircuit breakers, the reactance of the circuit breakerwhich is to interrupt the fault is not included incalculating fault current.

    (2) Above 600 amperes the reactance of molded casecircuit breakers are similar to those given in (a)

    For actual values, refer to manufacturers data.

    This table has been reprinted from IEEE Std 241-1990,

    IEEE Recommended Practice for Commercial Building

    Power Systems, copyright 1990 by the Institute of

    Electrical and Electronics Engineers, Inc. with the

    permission of the IEEE Standards Department.

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    Data Section

    "C" Values for Conductors and Busway

    Table 6. C Values for Conductors and BuswayCopper

    AWG Three Single Conductors Three-Conductor Cable

    or Conduit Conduit

    kcmil Steel Nonmagnetic Steel Nonmagnetic

    600V 5KV 15KV 600V 5KV 15KV 600V 5KV 15KV 600V 5KV 15KV

    14 389 389 389 389 389 389 389 389 389 389 389 389

    12 617 617 617 617 617 617 617 617 617 617 617 617

    10 981 981 981 981 981 981 981 981 981 981 981 981

    8 1557 1551 1557 1558 1555 1558 1559 1557 1559 1559 1558 1559

    6 2425 2406 2389 2430 2417 2406 2431 2424 2414 2433 2428 2420

    4 3806 3750 3695 3825 3789 3752 3830 3811 3778 3837 3823 3798

    3 4760 4760 4760 4802 4802 4802 4760 4790 4760 4802 4802 4802

    2 5906 5736 5574 6044 5926 5809 5989 5929 5827 6087 6022 5957

    1 7292 7029 6758 7493 7306 7108 7454 7364 7188 7579 7507 7364

    1/0 8924 8543 7973 9317 9033 8590 9209 9086 8707 9472 9372 9052

    2/0 10755 10061 9389 11423 10877 10318 11244 11045 10500 11703 11528 11052

    3/0 12843 11804 11021 13923 13048 12360 13656 13333 12613 14410 14118 13461

    4/0 15082 13605 12542 16673 15351 14347 16391 15890 14813 17482 17019 16012

    250 16483 14924 13643 18593 17120 15865 18310 17850 16465 19779 19352 18001

    300 18176 16292 14768 20867 18975 17408 20617 20051 18318 22524 21938 20163

    350 19703 17385 15678 22736 20526 18672 19557 21914 19821 22736 24126 21982

    400 20565 18235 16365 24296 21786 19731 24253 23371 21042 26915 26044 23517

    500 22185 19172 17492 26706 23277 21329 26980 25449 23125 30028 28712 25916

    600 22965 20567 47962 28033 25203 22097 28752 27974 24896 32236 31258 27766

    750 24136 21386 18888 28303 25430 22690 31050 30024 26932 32404 31338 28303

    1000 25278 22539 19923 31490 28083 24887 33864 32688 29320 37197 35748 31959

    Aluminum

    14 236 236 236 236 236 236 236 236 236 236 236 236

    12 375 375 375 375 375 375 375 375 375 375 375 375

    10 598 598 598 598 598 598 598 598 598 598 598 598

    8 951 950 951 951 950 951 951 951 951 951 951 951

    6 1480 1476 1472 1481 1478 1476 1481 1480 1478 1482 1481 1479

    4 2345 2332 2319 2350 2341 2333 2351 2347 2339 2353 2349 2344

    3 2948 2948 2948 2958 2958 2958 2948 2956 2948 2958 2958 2958

    2 3713 3669 3626 3729 3701 3672 3733 3719 3693 3739 3724 3709

    1 4645 4574 4497 4678 4631 4580 4686 4663 4617 4699 4681 4646

    1/0 5777 5669 5493 5838 5766 5645 5852 5820 5717 5875 5851 5771

    2/0 7186 6968 6733 7301 7152 6986 7327 7271 7109 7372 7328 7201

    3/0 8826 8466 8163 9110 8851 8627 9077 8980 8750 9242 9164 8977

    4/0 10740 10167 9700 11174 10749 10386 11184 11021 10642 11408 11277 10968

    250 12122 11460 10848 12862 12343 11847 12796 12636 12115 13236 13105 12661

    300 13909 13009 12192 14922 14182 13491 14916 14698 13973 15494 15299 14658

    350 15484 14280 13288 16812 15857 14954 15413 16490 15540 16812 17351 16500

    400 16670 15355 14188 18505 17321 16233 18461 18063 16921 19587 19243 18154

    500 18755 16827 15657 21390 19503 18314 21394 20606 19314 22987 22381 20978

    600 20093 18427 16484 23451 21718 19635 23633 23195 21348 25750 25243 23294

    750 21766 19685 17686 23491 21769 19976 26431 25789 23750 25682 25141 23491

    1000 23477 21235 19005 28778 26109 23482 29864 29049 26608 32938 31919 29135

    Note: These values are equal to one over the impedance per foot for impedances found in Table 5, Page 26.

    Ampacity Busway

    Plug-In Feeder High Impedance

    Copper Aluminum Copper Aluminum Copper

    225 28700 23000 18700 12000

    400 38900 34700 23900 21300

    600 41000 38300 36500 31300

    800 46100 57500 49300 44100

    1000 69400 89300 62900 56200 15600

    1200 94300 97100 76900 69900 16100

    1350 119000 104200 90100 84000 17500

    1600 129900 120500 101000 90900 19200

    2000 142900 135100 134200 125000 20400

    2500 143800 156300 180500 166700 21700

    3000 144900 175400 204100 188700 23800

    4000 277800 256400

    Note: These values are equal to one over the impedance per foot for

    impedances in Table 7, Page 28.

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    Data Section

    Busway Impedance Data

    Table 7. Busway Impedance Data (Ohms per 1000 Feet Line-to-Neutral, 60 Cycles)Plug-In Busway

    Copper Bus Bars Aluminum Bus Bars

    Ampere Rating Resistance Reactance Impedance Resistance Reactance Impedance

    225 0.0262 0.0229 0.0348 0.0398 0.0173 0.0434

    400 0.0136 0.0218 0.0257 0.0189 0.0216 0.0288

    600 0.0113 0.0216 0.0244 0.0179 0.0190 0.0261

    800 0.0105 0.0190 0.0217 0.0120 0.0126 0.0174

    1000 0.0071 0.0126 0.0144 0.0080 0.0080 0.0112

    1200 0.0055 0.0091 0.0106 0.0072 0.0074 0.0103

    1350 0.0040 0.0072 0.0084 0.0065 0.0070 0.0096

    1600 0.0036 0.0068 0.0077 0.0055 0.0062 0.0083

    2000 0.0033 0.0062 0.0070 0.0054 0.0049 0.0074

    2500 0.0032 0.0062 0.0070 0.0054 0.0034 0.0064

    3000 0.0031 0.0062 0.0069 0.0054 0.0018 0.0057

    4000 0.0030 0.0062 0.0069

    5000 0.0020 0.0039 0.0044

    Low-Impedance Feeder Busway

    225 0.0425 0.0323 0.0534 0.0767 0.0323 0.0832

    400 0.0291 0.0301 0.0419 0.0378 0.0280 0.0470

    600 0.0215 0.0170 0.0274 0.0305 0.0099 0.0320

    800 0.0178 0.0099 0.0203 0.0212 0.0081 0.0227

    1000 0.0136 0.0082 0.0159 0.0166 0.0065 0.0178

    1200 0.0110 0.0070 0.0130 0.0133 0.0053 0.0143

    1350 0.0090 0.0065 0.0111 0.0110 0.0045 0.0119

    1600 0.0083 0.0053 0.0099 0.0105 0.0034 0.0110

    2000 0.0067 0.0032 0.0074 0.0075 0.0031 0.0080

    2500 0.0045 0.0032 0.0055 0.0055 0.0023 0.0060

    3000 0.0041 0.0027 0.0049 0.0049 0.0020 0.0053

    4000 0.0030 0.0020 0.0036 0.0036 0.0015 0.0039

    5000 0.0023 0.0015 0.0027

    The above data represents values which are a composite of those obtained by a survey of industry; values tend to be on the low side.

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    Data Section

    Asymmetrical Factors

    Table 8. Asymmetrical FactorsRatio to Symmetrical RMS Amperes

    Short Circuit Short Maximum 1 phase Maximum 1 phase Average 3 phase

    Power Factor, Circuit Instantaneous RMS Amperes at RMS Amperes at

    Percent* X/R Ratio Peak Amperes Mp 1/2 Cycle Mm 1/2 Cycle Ma*

    (Asym.Factor)*

    0 2.828 1.732 1.394

    1 100.00 2.785 1.697 1.374

    2 49.993 2.743 1.662 1.354

    3 33.322 2.702 1.630 1.336

    4 24.979 2.663 1.599 1.318

    5 19.974 2.625 1.569 1.302

    6 16.623 2.589 1.540 1.286

    7 14.251 2.554 1.512 1.271

    8 13.460 2.520 1.486 1.256

    9 11.066 2.487 1.461 1.242

    10 9.9301 2.455 1.437 1.229

    11 9.0354 2.424 1.413 1.216

    12 8.2733 2.394 1.391 1.204

    13 7.6271 2.364 1.370 1.193

    14 7.0721 2.336 1.350 1.182

    15 6.5912 2.309 1.331 1.172

    16 6.1695 2.282 1.312 1.162

    17 5.7947 2.256 1.295 1.152

    18 5.4649 2.231 1.278 1.144

    19 5.16672 2.207 1.278 1.135

    20 4.8990 2.183 1.247 1.127

    21 4.6557 2.160 1.232 1.119

    22 4.4341 2.138 1.219 1.112

    23 4.2313 2.110 1.205 1.105

    24 4.0450 2.095 1.193 1.099

    25 3.8730 2.074 1.181 1.092

    26 3.7138 2.054 1.170 1.087

    27 3.5661 2.034 1.159 1.081

    28 3.4286 2.015 1.149 1.076

    29 3.3001 1.996 1.139 1.071

    30 3.1798 1.978 1.130 1.064

    31 3.0669 1.960 1.122 1.062

    32 2.9608 1.943 1.113 1.057

    33 2.8606 1.926 1.106 1.057

    34 2.7660 1.910 1.098 1.050

    35 2.6764 1.894 1.091 1.046

    36 2.5916 1.878 1.085 1.043

    37 2.5109 1.863 1.079 1.040

    38 2.4341 1.848 1.073 1.037

    39 2.3611 1.833 1.068 1.034

    40 2.2913 1.819 1.062 1.031

    41 2.2246 1.805 1.058 1.029

    42 2.1608 1.791 1.053 1.027

    43 2.0996 1.778 1.049 1.024

    44 2.0409 1.765 1.045 1.023

    45 1.9845 1.753 1.041 1.021

    46 1.9303 1.740 1.038 1.019

    47 1.8780 1.728 1.035 1.017

    48 1.8277 1.716 1.032 1.016

    49 1.7791 1.705 1.029 1.014

    50 1.7321 1.694 1.026 1.013

    55 1.5185 1.641 1.016 1.008

    60 1.3333 1.594 1.009 1.004

    65 1.1691 1.517 1.005 1.001

    70 1.0202 1.517 1.002 1.001

    75 0.8819 1.486 1.0008 1.0004

    80 0.7500 1.460 1.0002 1.0001

    85 0.6198 1.439 1.00004 1.00002

    100 0.0000 1.414 1.00000 1.00000

    *Reprinted by permission of National Electrical Manufacturer's Association from

    NEMA Publication AB-1, 1986, copyright 1986 by NEMA.

    Selective Coordination (Blackout Prevention)Having determined the faults that must be

    interrupted, the next step is to specify Protective

    Devices that will provide a Selectively Coordinated

    System with proper Interrupting Ratings.

    Such a system assures safety and reliability

    under all service conditions and prevents needless

    interruption of service on circuits other than the one

    on which a fault occurs.

    The topic of Selectivity will be Discussed in the

    next Handbook, EDP II.

    Component Protection (Equipment Damage Prevention)Proper protection of electrical equipment

    requires that fault current levels be known. The

    characteristics and let-through values of the

    overcurrent device must be known, and compared

    to the equipment withstand ratings. This topic ofComponent Protection is discussed in the third

    Handbook, EDP III.